Mathematics

Linear Algebra

510 Questions

Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.

Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties

Linear Algebra Questions

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta =\begin{vmatrix} a _{11} & a _{12} & a _{13}\ a _{21} & a _{22} & a _{23}\ a _{31} & a _{32} & a _{33} \end{vmatrix}$ and $c _{ij}=\left ( -1 \right )^{i+j}$ (determinant obtained by deleting ith row and jth column), then $\begin{vmatrix} c _{11} & c _{12} & c _{13}\ c _{21} & c _{22} & c _{23}\ c _{31} & c _{32} & c _{33} \end{vmatrix}=\Delta ^{2}$



If $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$ and $\Delta =\begin{vmatrix}
x^{3}-1 & 0 & x-x^{4}\
0 & x-x^{4} & x^{3}-1\
x-x^{4} & x^{3}-1 & 0
\end{vmatrix}$, then

  1. $\Delta =7$
  2. $\Delta =343$
  3. $\Delta =-49$
  4. $\Delta =49$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$
$\begin{vmatrix} c _{ 11 } & c _{ 12 } & c _{ 13 } \ c _{ 21 } & c _{ 22 } & c _{ 23 } \ c _{ 31 } & c _{ 32 } & c _{ 33 } \end{vmatrix}=\begin{vmatrix} x^{ 3 }-1 & 0 & x-x^{ 4 } \ 0 & x-x^{ 4 } & x^{ 3 }-1 \ x-x^{ 4 } & x^{ 3 }-1 & 0 \end{vmatrix}$
$\Delta ={ 7 }^{ 2 }=49$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

Let $\Delta _0=\begin{bmatrix}a _{11} & a _{12}  & a _{13}\a _{21}  & a _{22} &a _{23} \ a _{31} & a _{32} & a _{33}\end{bmatrix}$ (where $\Delta _0 \neq  0$) and let $\Delta _1$ denote the determinant formed by the cofactors of elements of $\Delta _0$ and $\Delta _2$ denote the determinant formed by the cofactor at $\Delta _1$ and so on $\Delta _n$ denotes the determinant formed by the cofactors at $\Delta _{n-1}$ then the determinant value of $\Delta _{n}$ is

  1. $\Delta _0^{2n}$
  2. $\Delta _0^{2^n}$
  3. $\Delta _0^{n^2}$
  4. $\Delta _0^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta _1=\Delta ^2 _0,\Delta _2=\Delta ^2 _1=\Delta ^{2^2} _0$
$\therefore \Delta _n=\Delta ^{2n} _0$

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x=cy+bz , y=az+cx$ and $z=bx+ay$, then ${a^2} + {b^2} + {c^2} + 2abc $ is equal to

  1. 2

  2. -1

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$a, b, c$ real numbers

$x, y, z$ real numbers not all zero

$x=cy+bz\rightarrow x-cy-bz=0 --- (1)$

$y=az+cx\rightarrow cx+y-az=0---(2)$

$z=bx+cy\rightarrow -bx-ay+z=0 --- (3)$

The system of equation have trivial solution then

$\begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix}=0$

$\Rightarrow 1(1-a^{2})+c(-c-ab)-b(a+b)=0$

$\Rightarrow a^{2}+b^{2}+c^{2}+2abc=1$

$D$ is correct
Multiple choice terms related to matrices matrices and determinants matrices algebra maths

 For what value of
k, the matrix $A = \begin{bmatrix} 4 & 3 -k\\ 1 & 2 \end{bmatrix}$ is
not invertible?

  1. (a)$k = -5,$
  2. (b)$k=5,$
  3. (c)$k=0$
  4. (d)$k=3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A matrix is not invertible if its determinant is zero. Det(A) = (4 * 2) - (1 * (3 - k)) = 8 - 3 + k = 5 + k. Setting 5 + k = 0 gives k = -5.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If the traces of $A, B$ are $20$ and $-8$, then the trace of $A+B$ is:

  1. $12$
  2. $-12$
  3. $28$
  4. $-28$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 the trace of an $n\times n$ square matrix A is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right) of A

the traces of A,B are 20 and −8, then the trace of $A+B$ is $trac(A+B)=trace(A)+trac(B)=20+-8=12$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If$A=\left[ \begin{matrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{matrix} \right] $ , then  trace of matrix $A$ is

  1. $17$
  2. $25$
  3. $3$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $A=\left[ \begin{matrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{matrix} \right] $
$tr(A)=sum\ of\ diagonal\ matrix$
$tr(A)=1+7+9=17$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $\displaystyle :A= \left [ a _{ij} \right ]$ is a scalar matrix of order $\displaystyle :n\times n$ such that $\displaystyle :a _{ij}= k $ for all then trace of A is equal to

  1. $\displaystyle \:nk$
  2. $\displaystyle \:n+k$
  3. $\displaystyle \:n/k$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By definition of trace of a scalar matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$

$=k+k+....k $ (upto n times)

$tr(A)=nk$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $A = \left[ {{a _{ij}}} \right]$ and ${a _{ij}} = i\left( {i + j} \right)$ then trace of $A=$

  1. $\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
  2. $\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{3}$
  3. $\frac{{n\left( {n + 1} \right)}}{2}$
  4. $\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $A=[a _{i,j}]$ and  $a _{i,j}=i(i+j)$ 

let the order of $A$ = $n\times n$

trace of $A$ =sum of all diagonal elements i.e.,$\sum[a _{i,j}]$ where $i=j$

$a _{1,1}=1(1+1)=2$
$a _{2,2}=2(2+2)=8$
$a _{3,3}=3(3+3)=18$
              $.$
              $.$
              $.$
$a _{n,n}=n(n+n)=2n^2$


$Trace$ $of$ $ A=$ $a _{1,1}+a _{2,2}+a _{3,3}+...........+a _{n,n}$ 
              $A=$  $2+8+18+.................+2n^2$
              $A=$  $2[1+4+9+................n^2]$

              $A=$  $2 \times [\frac{n(n+1)(2n+1)}{6}]$

              $A=$  $\frac{n(n+1)(2n+1)}{3}$

               $\therefore Opt$ $is$ $[B]$

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

Let $A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$ and $2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$, then $tr(A)-tr(B)$ has the value equal to

  1. 0

  2. 1

  3. 2

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$         .....(i)

$2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$         .....(ii)

$\Rightarrow 4A-2B=\begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix}$      ....(iii)

Adding (i) and (iii), we get
$5A=\begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$

$\Rightarrow A=\begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$
So, $tr(A)=1$

Now, by eq(ii),
$B=\begin{bmatrix} 2 & 0 & 4 \ 4 & -2 & 6 \ -2 & 2 & 2 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$

$\Rightarrow B=\begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$
So, $tr(B)=-1$
Now, $tr(A)-tr(B)=1+1=2$